Question: You have found the following ages (in years) of all 5 tigers at your local zoo: $ 22,\enspace 15,\enspace 14,\enspace 11,\enspace 19$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{22 + 15 + 14 + 11 + 19}{{5}} = {16.2\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $5.8$ years $33.64$ years $^2$ $15$ years $-1.2$ years $1.44$ years $^2$ $14$ years $-2.2$ years $4.84$ years $^2$ $11$ years $-5.2$ years $27.04$ years $^2$ $19$ years $2.8$ years $7.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{33.64} + {1.44} + {4.84} + {27.04} + {7.84}} {{5}} $ $ {\sigma^2} = \dfrac{{74.8}}{{5}} = {14.96\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{14.96\text{ years}^2}} = {3.9\text{ years}} $ The average tiger at the zoo is 16.2 years old. There is a standard deviation of 3.9 years.